Answer is: n<span>o, because the ion product is less than the Ksp of lead iodide. </span>
Chemical dissociation 1: KI(s) → K⁺(aq) + I⁻(aq).
Chemical dissociation 2: Pb(NO₃)₂(s) → Pb²⁺(aq) + 2NO₃⁻(aq).
Chemical reaction: Pb²⁺(aq) + 2I⁻(aq) → PbI₂(s).
Ksp(PbI₂) = 7.1·10⁻⁹.
V = 500 mL ÷ 1000 mL/L = 0.5 L.
c(KI) = c(I⁻) = 0.0025 mol ÷ 0.5 L.
c(I⁻) = 0.005 M.
c(Pb(NO₃)₂) = c(Pb²⁺) = 0.00004 mol ÷ 0.5 L.
c(Pb²⁺) = 0.00008 M.
Q = c(Pb²⁺) · c(I⁻)².
Q = 8·10⁻⁵ M · (5·10⁻³ M)².
Q = 2·10⁻⁹; <span> the ion product.</span>
15 is the group that phosphorus is found in.
The red is an oxygen molecule. 1 oxygen and 2 hydrogen =h2o or water!
Answer:
The fraction of water body necessary to keep the temperature constant is 0,0051.
Explanation:
Heat:
Q= heat (unknown)
m= mass (unknown)
Ce= especific heat (1 cal/g*°C)
ΔT= variation of temperature (2.75 °C)
Latent heat:
ΔE= latent heat
m= mass (unknown)
∝= mass fraction (unknown)
ΔHvap= enthalpy of vaporization (539.4 cal/g)
Since Q and E are equal, we can match both equations:

Mass fraction is:


∝=0,0051